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Problemas de ecuaciones exponenciales y logaritmos
Problemas de ecuaciones exponenciales y logaritmos
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Problemas de ecuaciones exponenciales y logaritmos

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  • 1. Problema 1 4x − 3x−1/2 = 3x+1/2 − 22x−1Ordenando 22x + 22x−1 = 3x+1/2 + 3x−1/2factorizando 22x (1 + 1/2) = 3x (31/2 + 1/31/2 ) 3 3+1 22x ( ) = 3x ( 1/2 ) 2 3 3 22 22x ( ) = 3x ( 1/2 ) 2 3ordenando y multiplicando en cruz tenemos 22x 23 = 1 +1 3x 32 3 22x 22 2 x = 3 3 32 2 2 22 3 ( )x = ( ) 2 3 3entonces como ambos lados de la ecuacion tienen la misma base obtenemos: 3 x= 2 Problema 2 1 1 1 3x+ 2 = 4x+ 2 − 3x− 2ordenando 1 1 1 3x+ 2 + 3x− 2 = 4x+ 2factorizando 3xen el lado izquierdo de la ecuacion tenemos: 1 1 1 3x (3 2 + 1 ) = 4x+ 2 32 3+1 1 3x ( 1 ) = 4x+ 2 3 2 1 x 3 4x+ 2 1 = 3 2 4 1
  • 2. ordenando y multiplicando en cruz 1 3x 4x 4 2 1 = 32 4 1 3x 4x 4 2 x = 4 4 3 x 3 1 ( ) = ( )2 4 4como ambos lados de la ecuacion tienen la misma base entonces 1 x= 2Problema 3 log4 {2 log3 [1 + log2 (1 + 3 log2 x)]} = 1 2 Solucion: Aplicando la definicion de logaritmo. 1 2 log3 [1 + log2 (1 + 3 log2 x)] = 4 22 log3 [1 + log2 (1 + 3 log2 x)] = 2simplificando el 2 log3 [1 + log2 (1 + 3 log2 x)] = 1Aplicando de nuevo la definicion de logaritmo1 + log2 (1 + 3 log2 x) = 3log2 (1 + 3 log2 x) = 2 de nuevo la definicion de lograitmo22 = 1 + 3 log2 x4 − 1 = 3 log2 x3 = 3 log2 x1 = log2 xY por ultimo aplicamos la definicion de logaritmo:x=2 2

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